# 12.15: Combination Problems

**At Grade**Created by: CK-12

**Practice**Combination Problems

### Let's Think About It

**License**: CC BY-NC 3.0

The ultimate Frisbee team just received their practice Frisbees. The Frisbees come in yellow, red, and blue. Marcos, the team captain, gives them out to the players two at a time. How many color combinations are possible?

In this concept, you will learn how to calculate possible combinations.

### Guidance

Once you have determined that you are using combinations, it is necessary to count the combinations.

There are several different ways to count combinations. When counting, try to keep the following in mind:

- Go one by one through the items. Don’t stop your list until you’ve covered every possible link of one item to all other items.
- Keep in mind that order doesn’t matter. For combinations, there no difference between \begin{align*}AB\end{align*}
AB and \begin{align*}BA\end{align*}BA . So if both \begin{align*}AB\end{align*}AB and \begin{align*}BA\end{align*}BA are on your list, cross one of the choices off your list. - Check your list for repeats. If you accidentally listed a combination more than once, cross the extra listings off your list.

James needs to choose a 2-color combination for his intramural team t-shirts. How many different 2-color combinations can James make out of red, blue, and yellow?

One way to find the number of combinations is to make a tree diagram. Here, if red is chosen as one color, that leaves only blue and yellow for the second color.

The diagram shows all 6 *permutations* of the 3 colors. But we are counting combinations here, so order doesn’t matter. Therefore, in this tree diagram, we will cross out all outcomes that are repeats. For example, the first red-blue is no different from blue-red, so we’ll cross out blue-red. In all, there are 3 combinations that are not repeats.

This method of making a tree diagram and crossing out repeats is reliable, but it is not the only way to find combinations.

Let's look at an example.

James has added a fourth color, green, from which to choose in selecting a 2-color combination for his intramural team. How many different 2-color combinations can James make out of red, blue, yellow, and green?

First, write the choices. Match the first choice, red, with the second, blue. Add the combination, red-blue, to your list. Match the other choices in turn. Add the combinations to your list.

Next, move to the second choice, blue. Match blue up with every possible partner. Add the combinations to your list.

Then, move to the third choice, yellow. There is only one combination left to match it with. Add the combination to your list.

The answer is there are 6 combinations.

Sometimes, you won’t want to use all of the possible options in the combination. Think about it as if you have 16 flavors of ice cream, but you only want to use three flavors at a time. With an example like this one, you are looking for combinations of object where only a certain number of them are used in any one combination. This happens a lot with teams.

Let's look at an example with teams.

How many different 2-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form?

First, start with Jean. Add all combinations that begin with Jean to your list.

\begin{align*}&\underline{\text{Combination}} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \underline{\text{List}}\\ &\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean}\\ &\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine}\\ &\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Lurleen}\\ &\text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Doreen}\end{align*}

Next, go through all combinations that begin with Dean, Francine, and Lurleen.

\begin{align*}&\underline{\text{Combination}} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \underline{\text{List}}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Francine}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Francine-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Francine-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Lurleen-Doreen}\end{align*}

The answer is there are 10 combinations.

Here is another example.

How many different 3-player soccer teams can Jean, Dean, Francine, Lurleen, and Doreen form?

First, start with Jean. Add all combinations that begin with Jean to your list.

Next, go through all combinations that begin with Dean, Francine, and Lurleen.

\begin{align*}&\underline{\text{Combination}} \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \underline{\text{List}}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean-Francine}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Dean-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Francine-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Jean-Lurleen-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean, Francine-Lurleen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Francine-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Dean-Lurleen-Doreen}\\ & \text{Jean, Dean, Francine, Lurleen, Doreen} \qquad \text{Francine-Lurleen-Doreen}\end{align*}

The answer is there are 10 combinations.

We can use a formula to help us to calculate combinations.

Suppose you have 5 marbles in a bag–red, blue, yellow, green, and white. You want to know how many combinations there are if you take 3 marbles out of the bag all at the same time. In combination notation, you write this as:

\begin{align*}{_5}C_3 \Longleftarrow 5 \ \text{items taken 3 at a time}\end{align*}

In general, combinations are written as:

\begin{align*}{_n}C_r \Longleftarrow n \ \text{items taken} \ r \ \text{at a time}\end{align*}

To compute \begin{align*}{_n}C_r\end{align*}

\begin{align*}{_n}C_r = \frac{n!}{r!(n - r)!}\end{align*}

where (n) is the number of objects and (r) is the number taken at any one time.

Now let’s look at applying the formula to the example.

For \begin{align*}{_5}C_3\end{align*}

\begin{align*}{_5}C_3 = \frac{5!}{3!(5 - 3)!} = \frac{5!}{3! 2!}\end{align*}

Next, simplify.

\begin{align*}{_5}C_3 = \frac{5 (4)(3)(2)(1)}{(3 \cdot 2 \cdot 1)(2 \cdot 1)} = \frac{120}{12} = 10\end{align*}

The answer is there are 10 possible combinations.

Here is another example.

Find \begin{align*}{_6}C_2\end{align*}

First, understand what \begin{align*}{_6}C_2\end{align*}

\begin{align*}{_6}C_2 \Longleftarrow 6 \ \text{items taken 2 at a time}\end{align*}

Next, set up the problem.

\begin{align*}{_6}C_2 = \frac{6!}{2!(6 -2)!}\end{align*}

Then, fill in the numbers and simplify.

\begin{align*}{_6}C_2 = \frac{6(5)(4)(3)(2)(1)}{(2 \cdot 1)(4 \cdot 3 \cdot 2 \cdot 1)} = \frac{720}{48} = 15\end{align*}

The answer is there are 15 possible combinations.

### Guided Practice

\begin{align*}{_5} C_4\end{align*}

First, understand what \begin{align*}{_5}C_4\end{align*}

\begin{align*}{_5}C_4 \Longleftarrow 5 \ \text{items taken 4 at a time}\end{align*}

Next, set up the problem.

\begin{align*}{_5}C_4 = \frac{5!}{4!(5 -4)!}\end{align*}

Then, fill in the numbers and simplify.

\begin{align*}{_5}C_4 = \frac{(5)(4)(3)(2)(1)}{(4 \cdot 3 \cdot 2 \cdot 1)(1)}= \frac{120}{24} = 5\end{align*}

The answer is there are 5 possible combinations.

### Examples

Find the number of combinations in each example.

#### Example 1

\begin{align*}{_5}C_2\end{align*}

First, understand what \begin{align*}{_5}C_2\end{align*}

\begin{align*}{_5}C_2 \Longleftarrow 5 \ \text{items taken 2 at a time}\end{align*}

Next, set up the problem.\begin{align*}{_5}C_2 = \frac{5!}{2!(5 -2)!}\end{align*}

Then, fill in the numbers and simplify.

\begin{align*}{_5}C_2 = \frac{(5)(4)(3)(2)(1)}{( 2 \cdot 1)(3 \cdot 2 \cdot 1)}= \frac{120}{12} = 10\end{align*}

The answer is there are 10 possible combinations.

#### Example 2

\begin{align*}{_4}C_3\end{align*}

First, understand what \begin{align*}{_4}C_3\end{align*}

\begin{align*}{_4}C_3 \Longleftarrow 4 \ \text{items taken 3 at a time}\end{align*}

Next, set up the problem.\begin{align*}{_4}C_3 = \frac{4!}{3!(4 -3)!}\end{align*}

Then, fill in the numbers and simplify.

\begin{align*}{_4}C_3 = \frac{(4)(3)(2)(1)}{(3 \cdot 2 \cdot 1)(1)}= \frac{24}{6} = 4\end{align*}

The answer is there are 4 possible combinations.

#### Example 3

\begin{align*}{_6}C_4\end{align*}

First, understand what \begin{align*}{_6}C_4\end{align*}

\begin{align*}{_6}C_4 \Longleftarrow 6 \ \text{items taken 4 at a time}\end{align*}

Next, set up the problem.

\begin{align*}{_6}C_4 = \frac{6!}{4!(6 -4)!}\end{align*}

Then, fill in the numbers and simplify.

\begin{align*}{_6}C_4 = \frac{(6)(5)(4)(3)(2)(1)}{(4 \cdot 3 \cdot 2 \cdot 1)(2 \cdot 1)}= \frac{720}{48} = 15\end{align*}

The answer is there are 15 possible combinations.

### Follow Up

**License**: CC BY-NC 3.0

Remember Marcos handing out Frisbees to his teammates?

He is handing them out 2 at a time and there are three colors: yellow, red, and blue. How many color combinations are possible?

First, understand what \begin{align*}{_3}C_2\end{align*} means.

\begin{align*}{_3}C_2 \Longleftarrow 3 \ \text{items taken 2 at a time}\end{align*}

Next, set up the problem.

\begin{align*}{_3}C_2 = \frac{3!}{2!(3 -2)!}\end{align*}

Then, fill in the numbers and simplify.

\begin{align*}{_3}C_2 = \frac{(3)(2)(1)}{(2 \cdot 1)( 1)} = \frac{6}{2} = 3\end{align*}

The answer is there are 3 possible color combinations of Frisbees.

### Video Review

### Explore More

Evaluate each factorial.

1. 5!

2. 4!

3. 3!

4. 8!

5. 9!

6. 6!

Evaluate each combination using combination notation.

7. \begin{align*}{_7} C_2\end{align*}

8. \begin{align*}{_7} C_6\end{align*}

9. \begin{align*}{_8} C_4\end{align*}

10. \begin{align*}{_9} C_6\end{align*}

11. \begin{align*}{_8} C_3\end{align*}

12. \begin{align*}{_{10}}C_7 \end{align*}

13. \begin{align*}{_{12}}C_9\end{align*}

14. \begin{align*}{_{11}}C_9\end{align*}

15. \begin{align*}{_{16}}C_{14}\end{align*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 12.15.

combination

Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.combination notation

Combination notation has the forms nCr and c(n, r) where n is the number of different units to choose from and r is the number of units in each group.combinatorics

Combinatorics is the study of permutations and combinations.n value

When calculating permutations with the TI calculator, the n value is the number of objects from which you are choosing.Permutation

A permutation is an arrangement of objects where order is important.permutation notation

Permutation notation is the form nPr or P(n, r), and indicates the number of ways that n objects can be ordered into groups of r items each.TI-84

The TI-84 calculator is a graphing calculator produced by Texas Instruments and is considered an “industry standard” for more advanced calculations.### Image Attributions

**[1]****^**License: CC BY-NC 3.0**[2]****^**License: CC BY-NC 3.0

## Description

## Learning Objectives

In this concept, you will learn how to calculate possible combinations.

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Dec 21, 2012## Last Modified:

Jan 26, 2016## Vocabulary

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